Solve the following equations for \( x \). \( \begin{array}{ll}17 .) 5^{2 x}=5^{2} & 18 \\ \text { 19. }(2.5)^{2 x+1}=(2.5)^{5} & 20 \\ \text { 21. } 10^{1-x}=100 & 22 \\ \text { 23. } 3(2.7)^{5 x}=8.1 & 24 \\ \text { 25. }\left(2^{x+1} \cdot 2^{-3}\right)^{2}=2 & 26\end{array} \)
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Let's solve these equations one by one: 17. \( 5^{2x} = 5^2 \) Since the bases are the same, we can set the exponents equal to each other: \( 2x = 2 \) Dividing both sides by 2 gives us: \( x = 1 \) --- 18. \( (2.5)^{2x+1} = (2.5)^5 \) Again, since the bases are the same, we can set the exponents equal: \( 2x + 1 = 5 \) Subtracting 1 from both sides gives us: \( 2x = 4 \) Dividing by 2 results in: \( x = 2 \) --- 19. \( 10^{1-x} = 100 \) We can rewrite 100 as \( 10^2 \): \( 10^{1-x} = 10^2 \) Setting the exponents equal gives us: \( 1 - x = 2 \) Solving for \( x \): \( -x = 2 - 1 \) So, \( -x = 1 \) or \( x = -1 \). --- 20. \( 3(2.7)^{5x} = 8.1 \) First, we can divide both sides by 3: \( (2.7)^{5x} = \frac{8.1}{3} \) Calculating \( \frac{8.1}{3} \): \( (2.7)^{5x} = 2.7 \) Since the bases are the same, we can equate the exponents: \( 5x = 1 \) Dividing both sides by 5, we find: \( x = \frac{1}{5} \) --- 21. \( (2^{x+1} \cdot 2^{-3})^{2} = 2 \) Simplifying the left side, we can use exponent rules: \( 2^{(x+1) - 3} = 2^{x - 2} \) Now raising it to the power of 2: \( (2^{x - 2})^{2} = 2^{2} \) We have: \( 2^{2(x - 2)} = 2^{2} \) Equating the exponents gives us: \( 2(x - 2) = 2 \) Dividing by 2 results in: \( x - 2 = 1 \) So, \( x = 3 \). --- In summary, the solutions are: 17. \( x = 1 \) 18. \( x = 2 \) 19. \( x = -1 \) 20. \( x = \frac{1}{5} \) 21. \( x = 3 \)